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# GATE’22 CE:- Daily Practice Quiz 30-Oct-2021

IIT Kharagpur has released the GATE 2022 exam dates. As per the GATE 2022 exam date, the exam will be conducted on February 5, 6, 12 & 13. Candidates preparing for GATE 2022 must attempt this Civil Engineering Quiz to boost your GATE exam preparation.

Quiz: Civil Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 marks
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. At the limit state of collapse, an RC beam is subjected to flexural moment 200 kN-m, shear force 20 kN and torque 9 kN-m. the beam is 300 mm wide and has a gross depth of 425 mm, with an effective cover of 25 mm. the equivalent nominal shear stress (τ_ve) as calculated by using the design code turns out to be lesser than the design shear strength (τ_c ) of the concrete. The equivalent flexural moment (M_eq ) for designing the longitudinal tension steel is
(a) 187 kN-m
(b) 200 kN-m
(c) 209 kN-m
(d) 213 kN-m

Q2. A fine-grained soil has liquid limit of 60 and plastic limit of 20. As per the plasticity chart, according to IS Classification, the soil is represented by the letter symbols
(a) CL
(b) CI
(c) CH
(d) CL-ML

Q3. In an iceberg, 15% of the volume projects above the sea surface. If the specific weight of sea water is 10.5 kN/m³, the specific weight of iceberg in kN/m³ is
(a) 12.52
(b) 9.83
(c) 8.93
(d) 6.83

Q4. The simply supported beam as shown in the figure. Find reaction at support A?
(a) 18 kN
(b) 22 kN
(c) 28 kN
(d) 30 kN

Q5. A lighthouse is visible just above the horizon at a certain station at the sea level. The distance between the station and the lighthouse is 40 km. the height of the lighthouse is approximately
(a) 187 m
(b) 137.7 m
(c) 107.7 m
(d) 87.3 m

Q6. Match List-I with List-II and select the correct answer:

 List-I List-II A.      Serviceability B.      Shear key C.      Shrinkage D.     Concrete spalling 1.      Sliding 2.      Deflection 3.      Cracking 4.      Corrosion

Codes:

A          B          C          D

(a)        1          3          4          2

(b)        2          1          3          4

(c)        1          3          2          4

(d)        2          1          4          3

Q7. A mild steel bar of square cross-section 40 mm × 40 mm is 400 mm long it is subjected to a longitudinal tensile stress of 440 N/mm² and lateral compressive stress is 200 N/mm² in perpendicular directions. E = 2 × 10^5 N/mm², μ = 0.3. what is the approximate elongation of the bar in the longitudinal direction?
(a) 0.44 mm
(b) 0.88 mm
(c) 0.22 mm
(d) 1 mm

Q8. For the plane frame as shown in the figure The degree of kinematic indeterminacy, neglecting axial deformation, is
(a) 3
(b) 5
(c) 7
(d) 9

Q9. An embankment is to be constructed with a granular soil (bulk unit weight =20 kN/m^3) on a saturated clayey silt deposit (undrained shear strength = 25 kPa). Assuming undrained general shear failure and bearing capacity factor of 5.7. the maximum height (in m) of the embankment at the point of failure is
(a) 7.1
(b) 5.0
(c) 4.5
(d) 2.5

Q10. A road is having a horizontal curve of 400 m radius on which a super-elevation of 0.07 is provided. The coefficient of lateral friction mobilized on the curve when a vehicle is travelling at 100 kmph is
(a) 0.07
(b) 0.13
(c) 0.15
(d) 0.4

Solutions

S1. Ans.(d)
Sol. Given,
M = 200 kN – m
T = 9 kN-m
D = 425 mm
B = 300 mm
Meq = ?
Meq = M +T/1.7 (1+D/B)

=200+9/1.7 (1+425/300)
=212.79≈213 kN-m

S2. Ans.(c)
Sol. w_L=60
w_P=20
Plasticity index (I_P)=w_L -w_P
= 60 – 20
= 40
Equation of A – line →
(I_P )_(A-line )=0.73 (W_L-20 )
= 0.73 (60 – 20)
= 29.2
(I_P )>(I_P )_(A-line), thus, the given soil lies above A – line & w_L is more than 50, hence soil will be CH.

S3. Ans.(c)
Sol. Specific weight of ice-berg = (1-0.15) × 10.5
= 0.85 × 10.5
= 8.93 kN/m³

S4. Ans.(b)
Sol. R_A+R_B=40 kN
Taking moment about point A →
ΣM_A=0
R_B×10-4×8-12×7-6×5-8×3-10×1=0
10R_B-32-84-30-24-10=0
10 R_B=180
▭(R_B=18 kN)
Reaction at A = (R_A ) = 40 – R_B
= 40 – 18
▭(R_A=22 kN)

S5. Ans.(c)
Sol. Given, d = 40 km
Height of lighthouse (h) will be –
h=0.0673 d^2
=0.0673×(40)^2
=107.68 m.

S6. Ans.(b)
Sol. Serviceability → Deflection
Shear key → Sliding
Shrinkage → Cracking
Concrete spalling → corrosion
→ Concrete spalling is caused by corrosion of steel reinforcement bars.

S7. Ans.(d)
Sol. Given,
A = 40×40 mm^2
L=400 mm
σ_x=400 N\/mm^2
σ_y = -200 N\/mm^2
E=2×10^5 N\/mm^2
μ=0.3
∆L= ?
Strain in longitudinal direction (∈_x ) will be-
∈_x=σ_x/E-μ σ_y/E
∆L/L=1/E (σ_x-μ σ_y )
∆L/400= 1/(2×10^5 ) [(400-0.3×(-200)]
▭(∆L=1 mm.)

S8. Ans.(b)
Sol. No. of joints (j) = 4
No. of member (m) = 3
No. of external reaction (re) = 5
Degree of kinematic indeterminacy (D_k) = ?
D_k=3j-re-m-r_r where,(r_r=Addition DOF due to joint hinge)
=(3×4)-5-3+1
=12-7
▭(D_k=5)

S9. Ans.(a)
Sol. Given, y_b=20kN\/m^3
Shear strength =S=C=25kN\/m^2
N_c=5.7
D_f=?
q_u=CN_C
y D_f=25×5.7 (∵q_u=yD_f )
20×D_f=25×5.7
▭(D_f=7.125m.)

S10. Ans.(b)
Sol. Given,
V=100 kmph
e=0.07
R=400m
f= ?
R=V^2/127(e+f)
400=(100)^2/127(0.07+f)
0.07+f=0.196
▭(f=0.13)

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